If either $\vec a = \vec 0$ or $\vec b = \vec 0$, then $\vec a \times \vec b = \vec 0$. Is the converse true? Justify your answer with an example.

When $\vec a = \vec 0$,

then $|\vec a| = 0$

$\vec a \times \vec b = |\vec a||\vec b|\sin \theta \hat n = (0)|\vec b|\sin \theta \hat n = \vec 0$

where, $\theta$ is the angle between $\vec a$ and $\vec b$.

Similarly, when $\vec b = \vec 0$, then $\vec a \times \vec b = \vec 0$

Conversely: Let $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$ and $\vec b = \lambda {a_1}\hat i + \lambda {a_2}\hat j + \lambda {a_3}\hat k$

Clearly $\vec a$ and $\vec b$ are parallel $\Rightarrow \theta = 0$

When $|\vec a| \ne 0$ and $|\vec b| \ne 0$
But, $\vec a \times \vec b = \vec 0$ if $\sin \theta \hat n = 0$.

Hence, $\vec a \times \vec b = \vec 0$

even if $\vec a \ne \vec 0$ and $\vec b \ne \vec 0$.

Let $\vec a = 2\hat i - \hat j + \hat k$ and $\vec b = 4\hat i - 2\hat j + 2\hat k$

$\vec a \times \vec b = \left| {\begin{array}{cccccccccccccccccccc}{\hat i}&{\hat j}&{\hat k}\\2&{ - 1}&1\\4&{ - 2}&2\end{array}} \right| = 0 \Rightarrow \vec a \times \vec b = 0$. But, $\vec a \ne 0$ and $\vec b \ne 0$